6. GRAPH AND MAP COLOURING


 Lawrence Burke
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1 6. GRPH ND MP COLOURING 6.1. Graph Colouring Imagine the task of designing a school timetable. If we ignore the complications of having to find rooms and teachers for the classes we could propose the following very simple model. We suppose that we have a certain number of subjects taken by a certain number of students. We ll suppose that all subjects require just one class. In setting up the timetable we must ensure that if two subjects have at least one student in common we must allocate different times. So imagine that the subjects are represented by the vertices of a graph and that we have an edge between subjects that have one or more common students. If we colour the vertices in such a way that adjacent vertices have different colours we can arrange for all subjects that were coloured with the same colour to be held at the same time. We could simply colour each vertex with a different colour. This would correspond to timetabling each subject at a different time. But there aren t enough hours in the week and it s important to try to use as few timeslots, or as few colours, as possible. This is one application that would benefit from a solution to the problem of colouring graphs with as few colours as possible. n ncolouring of a graph is an assignment of colours 1, 2,, n to the vertices of a graph so that adjacent vertices have different colours. The chromatic number, µ(g), of a graph G, is the smallest n for which an ncolouring exists. xample 1: For this graph G, µ(g) = colouring 4colouring 3colouring No 2coloring exists. round the pentagon we d have to alternate colours, but because of the odd number of sides we d end up with two adjacent vertices with the same colour. So 3 is the minimum number of colours. xample 2: The chromatic number of the complete graph K n is n. Proof: Since every vertex is adjacent to every other we need to use n distinct colours. µ = 3 µ = 5 97
2 very graph that contains a subgraph with chromatic number n must itself have chromatic number at least n. So, for example, the chromatic number of any graph that contains a triangle must be at least 3. 2 if n is even xample 3: The chromatic number of an nsided polygon is: 3 if n is odd µ = 2 µ = 3 xample 4: The chromatic number of K m,n is 2. Proof: Colour the m vertices one colour and the n vertices the other colour. µ(k 4,3 ) = 2 Theorem 1: The chromatic number of a tree (connected graph with no cycles) is 2 (except where there s only one vertex). Proof: Choose a vertex V 0. ach vertex has a unique distance from V 0. Use one colour to colour those vertices whose distance from V 0 is even and the other colour for those vertices whose distance is odd. Clearly this is a valid 2colouring The Chromatic Polynomial of a Graph The chromatic polynomial of a graph G is the number of ways of colouring G with k colours (a polynomial in k). It is denoted by Γ(G)(k). The chromatic number of G is the smallest value of k such that Γ(G)(k) is positive. For simplicity, unless we need to specify the parameter k, we ll write Γ(G)(k) simply as Γ(G). Theorem 2: Γ(K n ) = k(k 1) (k n + 1). Proof: Order the vertices in some way. There are k colours that can be assigned to the first vertex. ny one of the remaining colours can be assigned to the second vertex, and so on. xample 5: Γ(K 3 ) = k(k 1)(k 2). k colours k 1 k 2 colours colours 98
3 tree is a connected graph in which there are no cycles. In a tree there s a unique path (sequence of edges) from any vertex to any other and so there s a unique distance (counted in terms of the number of edges) between any two vertices. ny vertex whose distance from a given one is greatest must clearly have degree 1. So trees will always have vertices of degree 1. Theorem 3: If G is a tree with n vertices then Γ(G) = k(k 1) n 1. Proof: We use induction on the number of vertices. It s clearly true for n = 1. Suppose the result is true for trees with n vertices and that we have one with n + 1 vertices. Choose any vertex of degree 1 and remove it, together with the associated edge. By induction the resulting tree can be coloured with k colours in k(k 1) n 1 ways. Reinstating the deleted vertex, it s adjacent to only one coloured vertex and so can be coloured in k 1 ways, giving a total number of colourings of k(k 1) n. k 1 k 1 k 1 k k 1 k 1 k 1 k 1 k 1 k 1 We now show how to compute the chromatic polynomial of a graph. The method is inductive. From a given graph G we produce two simpler graphs G and G = and we use their chromatic polynomials to obtain the chromatic polynomial of G. Let G be a graph. Select any two adjacent vertices and B. Let G be the same graph with the edge B deleted. Let G = denote the graph G with vertices and B identified. (This means they become a single vertex and any edge having either or B as an endpoint now has, instead, this combined vertex.) B B = B G G G = xample 6: B B = B G G G = = B B G G G = B 99
4 Theorem 4: Let G be a graph. Select any two adjacent vertices and B. Then Γ(G) = Γ(G ) Γ(G = ). Proof: If we kcolour G we re free to colour and B the same colour. But we don t have to. We could just as validly colour them different colours (provided this was consistent with the colourings at the other vertices). There are two types of colourings of G in terms of what happens to and B 1B, B are given the SM colour 2B, B are given DIFFRNT colours B B G ach of these colourings of G gives a valid colouring of G =. G ach of these colourings of G gives a valid colouring of G. G = G Hence Γ G (k) = Γ G= (k) + Γ G (k). The result now follows algebraically. xample 7: Γ( ) = Γ( ) Γ( ) = k(k 1) 3 k(k 1)(k 2) = k(k 1)[(k 1) 2 (k 2)] = k(k 1)(k 2 3k + 3). xample 8: Γ( ) = Γ( ) Γ( ) Now Γ( ) = Γ( ).(k 1) = k(k 1) 2 (k 2) since once the triangle has been coloured the remaining vertex can be coloured any colour, except the colour of the vertex to which it is attached. and Γ( ) = Γ( ) Γ( ) = k(k 1) 4 k(k 1)(k 2 3k + 3) from example 7. Hence Γ( ) = k(k 1) 4 k(k 1)(k 2 3k + 3) k(k 1) 2 (k 2) = k(k 1)[(k 1) 3 (k 2 3k + 3) (k 1)(k 2)] = k(k 1)[k 3 3k 2 + 3k 1 k 2 +3k 3 k 2 + 3k 2] = k(k 1)(k 3 5k 2 + 9k 6) = k(k 1)(k 2)(k 2 3k + 3). It follows that the chromatic number of is 3. This was pretty obvious without all that calculation, but for a much more complicated graph, where the chromatic number is not so obvious, this could be a useful technique. lso, it could form the basis for a computer program to compute the chromatic number of a graph. Theorem 5: If C n is a cycle with n vertices then Γ(C n ) = (k 1) n + ( 1) n (k 1). Proof: Induction on n. Γ(C 3 ) = Γ(K 3 ) = k(k 1)(k 2) so it holds for n = 3. Suppose that it holds for n and let G = C n
5 Γ( ) = Γ( ) Γ( ). Γ(C n+1 ) = k(k 1) n Γ(C n ) = k(k 1) n (k 1) n ( 1) n (k 1) = (k 1) n+1 + ( 1) n+1 (k 1). Hence it holds for n + 1 and so, by induction, for all n Map Colouring Graph colouring is a purely combinatorial exercise that doesn t involve any topology. However it s a useful leadin to map colouring. s with graph colouring, when colouring maps, we must colour adjacent objects with different colours. But in the case of map colouring the objects are the regions, or faces, in the map and adjacent means that they share a common boundary. djacent counties in a map of the world are coloured differently to make the borders clear. Though there s no need to use the minimum number of colours it s an interesting question to ask what is the minimum number of colours we can use? n ncolouring of a map on a surface is a way of assigning one of n labels (colours) to each face so that adjacent faces have different colours. map is ncolourable if an ncolouring of it exists. The chromatic number, µ(m), of a map, M, is the smallest n for which M is ncolourable. xample 9: The chromatic number of the following map is 2: 4colouring 3colouring 2colouring but no 1colouring exists xample 10: The chromatic number of the following map is 3: 3colouring but no 2colouring exists 101
6 where xample 11: The chromatic number of the following map is 4: 4colouring but no 3coloring exists The chromatic number, µ(s), of a surface, S, is the largest chromatic number for any map that can be drawn on it. Now there s no a priori reason why this couldn t be infinite. However we ll show that every surface has a finite chromatic number. This means that there s an upper bound to the chromatic numbers of maps on a given surface. Upper bounds are harder to arrive at than lower bounds. To find a lower bound for the chromatic number of a surface we only have to draw a map on the surface, and the chromatic number of the surface will have to be at least as big as the chromatic number of that map. For example, the fact that in example 9 we have a map on a disk with chromatic number 2 means that µ(disk) 2. xample 12: µ(disk) 4. xample 11 gives a map, M, on a disk with µ(m) = 4. Hence µ(disk) 4. If there existed another map on the disk that needs 5 colours, we d have µ(disk) 5. But in fact it can be shown that every map on a disk can be coloured validly with 4 colours so in fact µ(disk) = 4. This is the famous Four Colour Theorem, proved in Recall that the degree of a vertex is the number of edges at that vertex. With a map we define the degree of a face to be the number of edges surrounding that face. xample 13: a vertex of degree 5 a face of degree 6 Clearly holes in a surface don t affect its chromatic number. So the chromatic number of a disk (sphere with one hole) will be the same as that of a sphere, or a cylinder (sphere with two holes). If required to find the chromatic number of a surface therefore, the first step is to remove all holes that is replace the surface by the corresponding surface with no holes. This is exactly what we do for embedding questions. lso, like embedding, we remove vertices of degrees 1 and 2 from a map. When removing a vertex of degree 1 we also remove the corresponding edge. When removing a vertex of degree 2 we combine the two edges into a single edge. In what follows we assume that this has already been done and that the surfaces have no boundaries and the degrees of the vertices are at least 3. In the following theorem we make use of the average degree of the vertices and the average 2 degree of the faces. If v is the average degree of the vertices then v = is the number of V edges and V is the number of vertices. Why not? The answer is that each edge contributes to V 102
7 and = =. =... the degree of each of its two endpoints. If we chopped each edge in two, so that the half edges were associated with only one vertex, then we d have 2 of these half edges. 2 Similarly, if f is the average degree of the faces, then f = F. gain, each edge is shared by two faces. xample 14: Here we ve labelled the vertices and faces with their degrees. Note that the outside has to be considered too, because we re regarding this as a map on a sphere. The average degree of the vertices is thus 8 8. Note that = 13, so this is simply V The average degree of the faces is 7 7. Note that this is simply F Theorem 6: Suppose that M is a map (with all vertices of degree 3) on a surface S (with no holes), with uler characteristic χ. Let V, F and be the numbers of vertices, faces and edges for M and let v be the average degree of the vertices and let f be the average degree of the faces. Then: 2fχ 2vχ (f 2)(v 2) = 4 V 4 F Proof: In counting the total number of edges by adding the degrees of the vertices or faces, each edge gets counted twice (it joins two vertices) and each face gets counted twice (it is adjacent to two faces). Hence ff = vv = 2. f Thus V = v F, establishing the second equality of the theorem. 2 Now v = V 2(V + F χ) = V 2F 2χ = 2 + V V 2v 2χ = 2 + f V 2fχ Thus vf = 2f + 2v V so vf 2f 2v + 4 = (f 2)(v 2) 2fχ = 4 V 6.4. Polyhedra polyhedron is a region in 3dimensional space, homeomorphic to a sphere and bounded by flat sides each of which is a polygon. There s no topological reason why the faces have to be flat, or the edges straight so it s convenient to consider polyhedra as maps on a sphere. Theorem 1 can therefore be applied to polyhedra by taking χ =
8 < Theorem 7: The average number of edges per face for any polyhedron is less than 6. Proof: We use the usual symbols for maps. 4f Since v 3 we have f 2 (f 2)(v 2) = 4 V 4, giving f < 6. consequence of this theorem is that there s no polyhedron where every face is a hexagon. There are, however, polyhedra where every face is a triangle (f = 3) or a square (f = 4) or even a pentagon (f = 5). But that s as far as we can go. If we extend the idea of a polyhedron to maps on other surfaces we can make the following conclusion about toroidal polyhedra (those which are homeomorphic to a torus). Theorem 8: For a map on a torus we have: (f 2)(v 2) = 4. Proof: This follows from theorem 1 by putting χ = 0. xample 15: The following is a 3 3 array of cubes, with the centre one removed. It can be considered as a map on a torus. There are eight vertices of degree 3, 16 of degree 4 and eight of degree 5. The average degree of the vertices is therefore v = 4. Since each face is a square, the average degree of the faces is f = 4. consequence of theorem 6 is that for a regular polyhedron on a torus the faces must be hexagons, squares or triangles (since v 2 must be an integer and so f 2 must divide 4). 104
9 . so,.... and Heawood s Theorem Theorem 9: For a map with F faces on a surface of weight χ 0 the average face degree is at most 6(χ 0 2) 6 + F 2vχ Proof: (f 2)(v 2) = 4 F 4 2χ v Hence f 2 = v 2 F v 2 4 2χ = v 2 F v 2 2 Thus f = χ v 2 F 6 1 χ F because v 3. 6(χ 0 2) So f 6 + F χ Corollary: If χ 0 2 then f 2 Proof: Since f F 1 (a given face cannot border more than all the remaining faces) we have: 6(χ 0 2) f 6 + f + 1 f 2 5f + 6(1 χ 0 ) 0. The greater of the two zeros of this quadratic is χ Theorem 10: (HWOOD 1910) Let S be a surface with no boundaries and weight χ χ If χ 0 2 then µ(s) INT lso χ(sphere) 6 and χ(projective plane) χ Proof: Suppose χ 0 2 and let k = 2 let n be the integer part of k. We prove by induction on the number of faces that every map on S can be ncoloured. map with one face can be coloured with a single colour and since n 4 we can easily do it with n colours. Now consider the general case. Suppose we have a certain map with at least two faces and assume that any map with fewer faces can be ncoloured. Since f k 1 there exists a face whose degree is at most k 1 and its degree is at most n 1. Shrink that face to a point. ll edges having one of the vertices of the chosen face as an endpoint will now meet at the same vertex. By induction we can ncolour this new map. Now replace the face in the now coloured map. It remains to show that this face can be validly coloured. 105
10 = = Since it borders at most n 1 other faces, at most n 1 colours must be avoided if we are to colour this remaining face. With n colours available this can be done. If χ 0 = 1 or 2 we can repeat the above argument by taking n = 6. Heawood only gave an upper bound. But surprisingly there are only two surfaces for which this upper bound is bigger than it need be. For all other surfaces Heawood s upper bound can be shown to be the actual value. The exceptions are the Klein bottle and the sphere. Heawood says µ(kb) 7 but in fact µ(kb) = 6. nd Heawood says µ(sphere) 6 while in fact µ(sphere) = 4. If S is a surface with no boundaries and weight χ 0 then: INT χ if χ µ(s) =. 6 if S is the Klein bottle 4 if S is the sphere In many cases we ll verify the chromatic number by using Heawood to get an upper bound and by constructing a suitable map to get the same value as a lower bound. xample 16: µ(projective plane) = χ Proof: Heawood says that µ INT INT The following map on the projective plane requires 6 colours so µ 6. Hence µ = 6. a b b a Note that, because of the orientation of the edges, if we cross over the edge a from region 6 we enter region 1, but it appears upside down, relative to what it would appear if we reached region 1 by travelling through region
11 xample 17: µ(torus) = 7. Proof: Heawood says µ 7. The following map on the torus requires 7 colours and so µ = 7. a b b a ach face is adjacent to every other face. It s remarkable that we can readily establish the chromatic number of surfaces like the projective plane and the torus while something so simple as the sphere (or equivalently the disk) is much more elusive. We ll soon prove that an upper bound for the sphere is 5. But the proof that in fact 4 is an upper bound (and hence the actual chromatic number) is exceedingly complicated The 5Colour Theorem Theorem 11: µ(sphere) 5. In other words, every map on a sphere can be coloured using 5 colours. Proof: Suppose there s a map M on the sphere that isn t 5colourable. We may suppose that we have a minimal counterexample that is, that every map with fewer faces is 5colourable. Let f be the average number of edges per face. Since f < 6, M has a face F with at most 5 edges. Remove one of these edges and combine the two faces on either side into a single face. 107
12 This map has fewer faces than M and so is 5colourable. Having 5coloured this slightly smaller map, now replace the edge and leave F uncoloured. If there are less than 5 different colours around F then there s one available with which to colour F. But this contradicts our assumption that M is not 5colourable. Hence F must be adjacent to exactly 5 faces and these must be coloured using 5 distinct colours. Since there s no colour left that we can use for F we ll need to modify the existing colouring. Let the faces surrounding F be F 1, F 2, F 3, F 4 and F 5 (in order around the outside of F, either clockwise or anticlockwise). Let the corresponding colours be C 1, C 2, C 3, C 4 and C 5. Delete all faces except those coloured C 1 or C 3. This may well disconnect M into several components. separate components 108
13 If F 1 and F 3 are in different components of this new map (in other words if you can t go from F 1 to F 3 by sticking to faces coloured C 1 and C 3 ) we could swap colours C 1 and C 3 in one of these components so that now F 1 and F 3 have the same colour (and, of course, maintaining a valid colouring). But now F would be surrounded by only 4 colours and so it could be validly coloured, to give a 5colouring of M. Since our assumption is that M doesn t have a 5colouring we conclude that F 1 and F 3 must be in the same component. That is, we can only get from one to the other by passing through C 1 or C 3 coloured faces. Such a closed path must separate F 2 and F 4. That is, one of F 2 and F 4 is on the inside of this closed path and the other is on the outside. (Of course, being on a sphere we can t say unambiguously which is inside and which is outside, but that doesn t matter.) C 1 C 2 C 3 C 4 This means that if we were to delete from M all faces except those coloured C 2 and C 4 the faces F 2 and F 4 must be in separate components. C 5 So we could swap colours C 2 and C 4 in one of these components and not in the other so that F 2 and F 4 now have the same colour. s before, this frees up one of the 5 colours with which to colour F and so complete the 5colouring of M. gain this is a contradiction. So we can t escape a contradiction. It follows that our assumption that there exists a non 5 colourable map on the sphere must be false. Hence every map on the sphere can be 5coloured. 109
14 XRCISS FOR CHPTR 6 xercise 1: Find the chromatic number of the following graph: xercise 2: (i) Find the chromatic polynomial of the following graph and express it in factorised form. (ii) Use this polynomial to find the number of ways of colouring the graph in (i) with 2 colours? How many ways with 3 colours? xercise 3: Find the number of ways of colouring the vertices of the following graph with 3 colours so that adjacent vertices have different colours. xercise 4: Find the number of ways of colouring the vertices of a square with k colours so that adjacent vertices have different colours. xercise 5: Find the number of ways of colouring the vertices of this graph, with 6 colours available, so that adjacent vertices have different colours. xercise 6: (i) Draw a diagram for K 6, the complete graph on 6 vertices. Draw a square with identified edges that represents the Klein bottle. (ii) mbed K 6 in the Klein bottle so that edges don t cross. (iii) Use this to draw, on a separate diagram, a map on the Klein bottle with six regions, each adjacent to the other five. (Finish with a clear picture in which only the edges of the map, and not the original graph, are visible. Label the regions 1 to 6. (iv) Let µ be chromatic number of the Klein bottle. What does your answer to (iv) tell you about µ? (v) Heawood s formula gives an upper bound for µ. Calculate this upper bound. (vi) Combine your answers to (v) and (vi) to determine µ as closely as possible. (Only use what you have proved. Do not simply quote the known value of µ.) 110
15 xercise 7: Let S be the sum of 3 projective planes and let n be the chromatic number of S. (a) Find n. (b) Construct a polygon with identified edges, P, to represent S. (c) Draw an embedding of K n, the complete graph on n vertices, on S. (d) Hence, or otherwise draw a map on P that requires n colours. xercise 8: Find the chromatic numbers of the following surfaces: (i) the Möbius Band; (ii) a cylinder with 3 holes; (iii) a sum of 10 projective planes. [You may assume that Heawood s upper bound is the actual value for all surfaces with no boundaries except the sphere and the Klein Bottle.] xercise 9: (a) regulation soccer ball is made up of patches, each of which is either a regular pentagon or a regular hexagon. ach vertex is surrounded by three patches. The pentagons are coloured black and the hexagons are coloured white. Based on this information alone show that there must be 12 pentagons. (b) Consider the pattern on a soccer ball to be a map on a sphere. It is made up of 12 pentagons, as predicted by (a), and 20 hexagons. Find the number of vertices and edges of this graph. 111
16 SOLUTIONS FOR CHPTR 6 xercise 1: Since the graph contains a cycle of length 5 there is clearly no 2colouring. The following is a 3coluring The chromatic number of this graph is therefore 3. xercise 2: (i) Let G 1 = G 2 = and G 3 = Then Γ(G 1 ) = Γ(G 2 ) Γ(G 3 ). Let G 4 = and G 5 = Then Γ(G 2 ) = Γ(G 4 ) Γ(G 5 ). Clearly Γ(G 4 ) = k(k 1) 6 and Γ(G 5 ) = k(k 1) 4 (k 2). Hence Γ(G 2 ) = k(k 1) 6 k(k 1) 4 (k 2) = k(k 1) 4 (k 2 3k + 3). Let G 6 = and G 7 = 112
17 Then Γ(G 3 ) = Γ(G 6 ) Γ(G 7 ). Clearly Γ(G 6 ) = k(k 1) 4 (k 2) and Γ(G 7 ) = k(k 1) 2 (k 2) 2. Hence Γ(G 3 ) = k(k 1) 4 (k 2) k(k 1) 2 (k 2) 2 = k(k 1) 2 (k 2)(k 2 3k + 3). So Γ(G 1 ) = k(k 1) 4 (k 2 3k + 3) k(k 1) 2 (k 2)(k 2 3k + 3) = k(k 1) 2 (k 2 3k + 3) 2. (ii) There are 2 ways of colouring the graph with 2 colours and 108 ways of colouring it with 3 colours. xercise 3: Γ ( ) = Γ( ) Γ( ) Γ( ) = (k 1)[(k 1) 5 + 1] = = 66 when k = 3. Γ( ) = Γ( ) Γ( ) = k(k 1)(k 2)(k 1) 2 k(k 1)(k 2)(k 1) = k(k 1) 2 (k 2) 2 = 12 when k = 3. Hence Γ( ) = = 54 when k = 3. xercise 4: Γ= k(k 1) 3 k(k 1)(k 2) = k(k 1)(k 2 2k + 1 k + 2) = k(k 1)(k 2 3k + 3). xercise 5: Let the graph be called G and consider the diagonal. G is a square and its chromatic polynomial is (k 1) 4 + (k 1). G = is a linear graph with 3 vertices and 2 edges. Its chromatic polynomial is k(k 1) 2. Hence µ(g) = (k 1) 4 + (k 1) + k(k 1) 2. When k = 6, µ = 780. xercise 6: (i) a b b a (ii) K 6 embedded in the KB
18 so = (iii) Joining a point in the middle of each face to the midpoint of each of the surrounding edges we get a map with 6 faces, each of which is adjacent to the other 5. This map requires 6 colours (iv) We need 6 colours to colour this map on the KB so µ(kb) 6. (v) Heawood s formula gives µ(kb) INT µ(kb) 7. (vi) So µ(kb) = 6 or 7. (In fact it is 6 but we haven t shown that.) xercise 7: (a) The weight of 3P is 3. By Heawood s formula n = INT (b), (c) (d) To obtain a map with 7 regions, each adjacent to the other 6, we take a point inside each face of the map obtained in (b) and join it to a point along each edge. 114
19 + = = = Removing the original edges we obtain a map on 3P in which each of the 7 regions are adjacent to the other 6. This map needs 7 colours. xercise 8: (i) µ( Möbius Band) = µ(projective Plane) = INT 2 6. (ii) µ( cylinder with 3 holes) = µ(sphere with 5 holes) = µ(sphere) = (iii) µ(sum of 10 projective planes) = INT xercise 9: (a) Let there be p pentagons and h hexagons. If V, F and are the numbers of vertices, faces and edges respectively then: F = p + h 2 = 5p + 6h (each edge belongs to 2 faces) 3V = 5p + 6h (each vertex belongs to 3 faces) 5p + 6h 5p + 6h Since V + F = 2 we have p + h This reduces to p = 12. (b) Clearly F = 32 and h = 20. Hence = 90 and V =
20 116
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